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## Homework Statement

given this ODE with initial conditions y(1)=0

[tex] \[

(x + y^2 )dx - 2xydy = 0

\][/tex]

## Homework Equations

solving this ODE gives us

[tex]\[y = \sqrt {x\ln (x)} \][/tex]

as we can see this equation is true only for x>=1

in order to use the theorem on existence and uniqueness we isulate for y'=f(x,y)

[tex]\[y' = \frac{{(x + y^2 )}}{{2xy}}\][/tex]

and we can see that when y=0 the equation is not defined

## The Attempt at a Solution

my question is

1) if x>=1 does that mean that the bound for y is y>=0?

2)if it meaas that y>=0 then should i conclude that the theorem on existence and uniqueness does not apply here since the function is not continuous thus we cant say that the solution is unique? what does it mean

thanks for the help